How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
Given the balanced equation2C3H8O + 9O2 --> 6CO2 + 8H2OTo find the number of moles CO2 that will be produced from 0.33 mol C3H8O, we must convert from moles to moles (mol --> mol conversion).0.33 mol C3H8O * 6 molecules CO2 = 0.99 mol CO2---------- 2 molecules C3H8O
8 mol
5 moles RbNO3 (3 moles O2/2 moles RbNO3) = 7.5 moles oxygen gas produced
How many moles of CO2 are produced when 2.1 mol of C2H2 react?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
The balanced chemical equation for the reaction between H2 and NH3 is: 3H2 + N2 → 2NH3 From the equation, we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, when 1.2 moles of H2 react, we can calculate the moles of NH3 produced as: 1.2 mol H2 * (2 mol NH3 / 3 mol H2) = 0.8 mol NH3.
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
Na2CO32 * 2 = 4 moles sodium.===========================
0,3 moles of nitrogen reacted.
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
Given the balanced equation2C3H8O + 9O2 --> 6CO2 + 8H2OTo find the number of moles CO2 that will be produced from 0.33 mol C3H8O, we must convert from moles to moles (mol --> mol conversion).0.33 mol C3H8O * 6 molecules CO2 = 0.99 mol CO2---------- 2 molecules C3H8O
The balanced equation for the combustion of ethane (C2H6) is: 2C2H6 + 7O2 -> 4CO2 + 6H2O From the equation, for every 2 moles of ethane burned, 4 moles of CO2 are produced. Therefore, if 5.60 mol of ethane are burned, then (5.60 mol / 2 mol ethane) * 4 mol CO2 = 11.2 mol of CO2 are produced.
8 mol